1,Cho \(a^2+b^2+c^2+3=2\left(a+b+c\right)\) .Cmr: \(a=b=c=1\)
2,Cho \(\left(a+b+c\right)^2=3\left(ab+ac+bc\right)\) .Cmr: \(a=b=c\)
3,Cho \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=\left(a+b-2c\right)^2+\left(b+c-2a\right)^2+\left(c+a-2b\right)^2\) .Cmr: \(a=b=c\)
4,Cho a,b,c,d là các số khác 0 và:
\(\left(a+b+c+d\right)\left(a-b-c+d\right)=\left(a-b+c-d\right)\left(a+b-c-d\right)\) .Cmr: \(\dfrac{a}{c}=\dfrac{b}{d}\)
5,Cho \(x^2-y^2-z^2=0\) .Cmr: \(\left(5x-3y+4z\right)\left(5x-3y-4z\right)=\left(3x-5y\right)^2\)
HELP ME!mik cần gấp lắm rồi!Thank trước nhé!
4) Ta có : A=(a+b+c+d)(a-b-c+d)=(a-b+c-d)(a+b-c-d)
=> (a+d)2 - (b+c)2= (a-d)2 - (c-b)2
=> a2+ d2+ 2ad - b2- c2- 2bc=a2 + d2 - 2ad - c2-b2+2bc
Rút gọn ta được: 4ad = 4bc => ad = bc =>\(\dfrac{a}{c}=\dfrac{b}{d}\)
1) a2+b2+c2+3=2(a+b+c) =>(a-1)2+(b-1)2+(c-1)2=0
=> a-1=b-1=c-1=0 => a=b=c=1 =>đpcm
2) (a+b+c)2=3(ab+bc+ac) =>(a-b)2+(b-c)2+(c-a)2=0
=>a-b=b-c=c-a=0 =>a=b=c
